Question 273955
 two pipes working together drain a water holding tank in 6 hours.
 if used alone to empty the tank, one takes 2 hours longer than the other.
 how long does each take to empty the tank if used alone?
:
Let x = time for one pipe to drain it by itself
then
(x+2) = time for the other pipe to by itself
;
Let the completed job = 1, (a drained tank)
:
Each pipe does a fraction of the job, the two fractions add up to one.
{{{6/x}}} + {{{6/((x+2))}}} = 1
Multiply equation by x(x+2), results:
6(x+2) + 6x = x(x+2)
;
6x + 12 + 6x = x^2 + 2x
:
12x + 12 = x^2 + 2x
Arrange as a quadratic equation
0 = x^2 + 2x - 12x - 12 
:
x^2 - 10x - 12 = 0
We have to use the quadratic formula to solve this:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation a=1; b=-10; c=-12
{{{x = (-(-10) +- sqrt(-10^2-4*1*-12 ))/(2*1) }}}
:
{{{x = (10 +- sqrt(100 -(-48) ))/2 }}}
{{{x = (10 +- sqrt(100 + 48 ))/2 }}}
{{{x = (10 +- sqrt(148 ))/2 }}}
We only want the positive solution
{{{x = (10 + 12.1655)/2 }}}
x = {{{22.1655/2}}}
x = 11.08 hrs, for the 1st pipe to drain it alone
and
11.08 + 2 = 13.08 hrs for the 2nd pipe alone
;
:
Check solution
6/11.08 + 6/13.08 = 
.54 + .46 = 1; confirms our solutions